D. prove
Three girls Reshma, Salma and Mandip are
playing a game by standing on a circle of radius
5m drawn in a park. Reshma throws a ball to
Salma, Salma to Mandip. Mandip to Reshma. If
the distance between Reshma and Salma and
between Salma and Mandip is 6m each, what is
the distance between Reshma and Mandip?
20m is situated in ac
Answers
Answer:
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Three girls Reshma, Salma and Mandeep are playing a game by standing on a circle of radius 5 m drawn in a park. Reshma throws a ball to Salma, Salma to Mandeep, Mandeep to Reshma. If the distance between Reshma and Salma and between Salma and Mandeep is 6 m each, what is the distance between Reshma and Mandeep?
Let R, S and M represent the position of Reshma, Salma and
Mandeep respectively.
Clearly △RSM is an isosceles triangle as
RS = SM = 6m
Join OS which intersect RM at A.
In △ROS and △MOS
OR = OM (Radii of the same circle)
OS = OS (Common)
RS = SM (Each 6cm)
∴ △ROS ≅ △MOS (By SSS congruence criterion)
∴ ∠RSO = ∠MSO (CPCT)
In △RAS and △MAS
AS = AS (Common)
∴ ∠RSA = ∠MSA (Since, ∠RSO = ∠MSO)
RS = MS (Given)
∴ △RAS ≅ △MAS (By SAS congruence criterion)
∴ ∠RAS = ∠MAS (CPCT)
Since, ∠RAS + ∠MAS = 180° (Linear pair)
⇒ ∠RAS = ∠MAS = 90°
Let OA = x m ⇒ AS = (5 - x)m
In right triangle RAS,
RS2 = RA2 + AS2
⇒62 = RA2 + (5 - x)2
⇒RA2 = 62 - (5 - x)2 ....(i)
In right triangle RAO,
RO2 = RA2 + OA2
⇒52 = RA2 + x2
⇒RA2 = 52 - x2 ...(ii)
From equation (i) and (ii), we get
62 - (5-x)2 = 52 - x2
62 - 52 = (5-x)2 - x2
36 - 25 = 25 + x2 - 10x - x2
11 = 25 - 10x ⇒ 10x = 14 ⇒ x = 1.4m
From equation (ii), we have
RA2 = 52 -(1.4)2 = 25 - 1.96
RA2 = 23.04 ⇒ RA = √23.04
RA = 4.8m
As the perpendicular from the centre of a circle bisects the chord.
∴ RM = 2RA
RM = 2 x 4.8 = 9.6m
Hence, distance between Reshma and Mandeep is 9.6m.
